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**[SOLVED] Area inside Polar Curves**

## Homework Statement

I have spent several hours beating myself up over this and I just can't seem to solve it. It's the only problem I haven't gotten correct and it is particularly frustrating. Can you math gods here save me? :)

Find the area of the region that lies inside both curves

[tex]r = 6 \sin (2 \theta) , \quad r = 6 \sin (\theta) [/tex]

## Homework Equations

Area inside a single polar curve: [tex]\frac{1}{2}\int{f(\theta)^{2} d\theta}[/tex]

## The Attempt at a Solution

They intersect at [tex]\frac{\pi}{3}[/tex] and 0

So, I take the area of the second equation from 0 to pi/3, and the area of the first equation from pi/3 to pi/2 since pi/2 is where it meets back at the origin.

[tex]\frac{1}{2}\int_{0}^{\pi/3}{(6\sin(\theta))^{2} d\theta} + \frac{1}{2}\int_{pi/3}^{\pi/2}{(6\sin(2\theta))^{2} d\theta} = \frac{-9(3\sqrt(3)-4\pi)}{8} \approx 8.2915[/tex]

The website (WebWork) that we input the answer to says it is not correct though, and I cannot figure out any other way that would make sense.